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Nitella [24]
3 years ago
8

A pendulum of mass 5.0 ???????? hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal fo

rce of 30.0 ????. What is the maximum angle of displacement of the swinging pendulum? How long does the pendulum have to be to have a period of 5.0 ????????co?????????????
Physics
1 answer:
Rudiy273 years ago
4 0

Answer:

a. 37.75°

b. 6.21 m

Explanation:

a. The horizontal force acting on a pendulum bob is given as:

F = mgsinθ

where m = mass of bob

g = acceleration due to gravity

θ = angle string makes with the vertical or angle of displacement

Making θ subject of formula, we have:

θ = sin^{-1}(\frac{F}{mg} )

θ = sin^{-1} (\frac{30}{5*9.8})

θ = 37.75°

The maximum angle of displacement is 37.75°

b. Period of a pendulum is given as:

T = 2\pi\sqrt{ \frac{L}{g} }

where L = length of string

Therefore, making L subject of formula:

L = \frac{gT^2}{4\pi^2}

L = \frac{9.8 * 5^2}{4\pi^2} \\\\\\L = 6.21 m

The string holding the pendulum has to be 6.21 m long.

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Explanation:

Given that,

A student covered a distance of 210 meters in 35 seconds.

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Speed, v = distance (d)/time (t)

So,

v=\dfrac{210\ m}{35\ s}\\\\=6\ m/s

We know that, 1 minute = 60 seconds

6\dfrac{m}{s}=6\times \dfrac{m}{(\dfrac{1}{60})\ \text{minutes}}\\\\=360\ \text{meters/minutes}

Hence, the student's speed is 6 m/s or 360 meters/minute.

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Explanation:

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3 years ago
Can someone help me wit hthis science question?
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see explanation  

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6 0
3 years ago
Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par
taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ =  5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

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The distance traveled is given by;

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Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0
3 years ago
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